IB DP Chemistry - R3.2.14 Gibbs free energy and cell potentialr - Study Notes - New Syllabus - 2026, 2027 & 2028
IB DP Chemistry – R3.2.14 Gibbs free energy and cell potential – Study Notes – New Syllabus
IITian Academy excellent Introduction to the Proton transfer reactions – Study Notes and effective strategies will help you prepare for your IB DP Chemistry exam.
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Reactivity 3.2.14 — Relationship Between \( \Delta G^\circ \) and Standard Cell Potential \( E^\circ_{\text{cell}} \)
Reactivity 3.2.14 — Relationship Between \( \Delta G^\circ \) and Standard Cell Potential \( E^\circ_{\text{cell}} \)
The spontaneity of a redox reaction under standard conditions can be predicted using either thermodynamic data (from enthalpy and entropy changes) or electrochemical data (from standard cell potentials).
Equation:
\( \Delta G^\circ = -nFE^\circ_{\text{cell}} \)
Where:
- \( \Delta G^\circ \) = standard Gibbs free energy change (J mol−1)
- \( n \) = number of moles of electrons transferred in the redox reaction
- \( F \) = Faraday constant = \( 9.65 \times 10^4\ \text{C mol}^{-1} \)
- \( E^\circ_{\text{cell}} \) = standard cell potential (V)
Interpretation of Sign:
- \( E^\circ_{\text{cell}} > 0 \Rightarrow \Delta G^\circ < 0 \) → reaction is spontaneous
- \( E^\circ_{\text{cell}} < 0 \Rightarrow \Delta G^\circ > 0 \) → reaction is non-spontaneous
Alternative Thermodynamic Approach:
Spontaneity can also be predicted using thermodynamic quantities such as enthalpy change \( \Delta H^\circ \) and entropy change \( \Delta S^\circ \) from:
\( \Delta G^\circ = \Delta H^\circ – T \Delta S^\circ \)
- If \( \Delta G^\circ < 0 \) → spontaneous
- If \( \Delta G^\circ > 0 \) → non-spontaneous
Summary of Spontaneity Predictions:
| Approach | Equation | Spontaneity Condition |
|---|---|---|
| Electrochemical | \( \Delta G^\circ = -nFE^\circ_{\text{cell}} \) | Spontaneous if \( E^\circ_{\text{cell}} > 0 \) |
| Thermodynamic | \( \Delta G^\circ = \Delta H^\circ – T \Delta S^\circ \) | Spontaneous if \( \Delta G^\circ < 0 \) |
Example
The cell reaction below occurs under standard conditions:
\( \text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s) \)
Given: \( E^\circ_{\text{cell}} = +1.10\ \text{V} \), and 2 moles of electrons are transferred.
Calculate \( \Delta G^\circ \) in kJ mol−1.
▶️Answer/Explanation
Use the equation:
\( \Delta G^\circ = -nFE^\circ_{\text{cell}} \)
\( = -(2)(9.65 \times 10^4)(1.10) \)
\( = -2.123 \times 10^5\ \text{J mol}^{-1} \)
\( = -212.3\ \text{kJ mol}^{-1} \)
The negative value confirms the reaction is spontaneous.
Example
For the redox reaction:
\( \text{Ag}^{+}(aq) + \text{e}^- \rightarrow \text{Ag}(s) \quad E^\circ = +0.80\ \text{V} \)
\( \text{Fe}^{2+}(aq) \rightarrow \text{Fe}^{3+}(aq) + \text{e}^- \quad E^\circ = -0.77\ \text{V} \)
Determine the standard Gibbs free energy change, \( \Delta G^\circ \), in kJ mol−1 for the overall reaction.
▶️Answer/Explanation
The overall reaction is:
\( \text{Fe}^{2+} + \text{Ag}^+ \rightarrow \text{Fe}^{3+} + \text{Ag} \)
\( E^\circ_{\text{cell}} = +0.80 – (-0.77) = +1.57\ \text{V} \)
\( n = 1 \) mole of electrons
\( \Delta G^\circ = -(1)(9.65 \times 10^4)(1.57) = -1.515 \times 10^5\ \text{J mol}^{-1} = -151.5\ \text{kJ mol}^{-1} \)
Spontaneous since \( \Delta G^\circ < 0 \).
Example
Calculate the standard Gibbs free energy change, \( \Delta G^\circ \), in kJ mol−1 for the reaction:
\( 3\text{Cu}^{2+}(aq) + 2\text{Al}(s) \rightarrow 3\text{Cu}(s) + 2\text{Al}^{3+}(aq) \)
Use the following standard reduction potentials:
\( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}(s) \quad E^\circ = +0.34\ \text{V} \)
\( \text{Al}^{3+} + 3e^- \rightarrow \text{Al}(s) \quad E^\circ = -1.66\ \text{V} \)
▶️Answer/Explanation
Oxidation (anode): \( \text{Al}(s) \rightarrow \text{Al}^{3+} + 3e^- \quad E^\circ = +1.66\ \text{V} \) (reverse of reduction)
Reduction (cathode): \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}(s) \quad E^\circ = +0.34\ \text{V} \)
So, \( E^\circ_{\text{cell}} = 1.66 + 0.34 = 2.00\ \text{V} \)
To balance electron transfer:
- \( \text{Cu}^{2+} \) needs 2e− per ion × 3 = 6 electrons total
- \( \text{Al} \) loses 3e− per atom × 2 = 6 electrons total
So \( n = 6 \)
Calculate \( \Delta G^\circ \)
\( \Delta G^\circ = -nFE^\circ_{\text{cell}} \)
\( = -(6)(9.65 \times 10^4)(2.00) \)
\( = -1.158 \times 10^6\ \text{J mol}^{-1} = -1158\ \text{kJ mol}^{-1} \)
The large negative \( \Delta G^\circ \) value confirms this reaction is highly spontaneous.